Question

A^x=b^y=c^z=d^u and a,b,c,d are in g.p then prove that x,y,z,u are in h.p

Answer 1

a^x = b^y = c^z = d^u
and a , b , c , d are in GP .

we can handle it in two parts

first part :-
a^x = b^y = c^z = K
a = K^{1/x}, b = K^{1/y}, c = K^{1/z}
where, a, b, c are in GP.
then,
b² = ac
K^{2/y} = K^{1/x}.K^{1/z}
K^{2/y} = K^{(x+z)/xz}
2/y = (x + y)/xz
y = 2xz/(x + y)

hence, 1/x , 1/y, 1/z are in AP
so, x , y, z are in HP

similarly do with second part :
b^y = c^z = d^u = K
b = K^{1/y} , c = K^{1/z}, d = K^{1/u}
where, b, c , d are in GP
so, c² = bd
and then putting above values
and we get,
z = 2yu/(u + y)

hence, 1/y, 1/z 1/u are in AP
so, y, z , u are in HP


hence,
x, y,z , u are in HP